Explanation Given: The mass of 8 pounds is attached to the 4-foot spring. The length of the spring after the mass is attached is 8 feet. Calculation: Let the displacement be s . The spring is 4 feet and length of the spring after the mass is attached is 8 feet. Thus, the elongation is 4 feet. According to Hooke’s law, m g = k s (1) Here, m is mass, g is gravity, k is spring constant and s is displacement. Substitute s = 4 and m g = 8 in the equation (1) as follows: 8 = k × 4 k = 8 4 k = 2 Therefore, the value of k is 2. Express the mass as, m = w g (2) Substitute w = 8 and g = 32 in the equation (2) as follows: m = 8 pound 32 ft / sec 2 = 1 4 slugs Thus, the mass is 1 4 slugs . The medium imparts a damping force equal to 2 times the instantaneous velocity. So, β = 2 . The general equation of motion for a spring mass system is, m x ″ + β x ′ + k x = 0 (3) Substitute the value of m = 1 4 , k = 2 , and β = 2 in the equation (3) as follows: 1 4 x ″ + 2 x ′ + 2 x = 0 Multiply the above equation by 4. 4 4 x ″ + 4 2 x ′ + 4 ⋅ 2 x = 0 x ″ + 4 2 x ′ + 8 x = 0 The auxiliary of above equation is, m 2 + 4 2 m + 8 = 0 By quadratic formula m is calculated as, m = − 4 2 ± ( 4 2 ) 2 − 4 × 1 × 8 2 = − 4 2 ± 32 − 32 2 = − 4 2 2 = − 2 2 Note that, it is a repeated root. The solution of differential equation with repeated roots is, x c = c 1 e m t + c 2 t e m t (4) Substitute the value of m = − 2 2 in the equation (4) as follows: x c = c 1 e − 2 2 t + c 2 t e − 2 2 t (5) The mass is released from the equilibrium position. So, x ( 0 ) = 0 . After differentiating with respect to t , it is represented as x ′ ( 0 ) = 5

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ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 3.8, Problem 26E

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The equation of motion and the time at which the mass attains its extreme displacement from equilibrium position and find the position of the mass at this instant.

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Chapter 3.8, Problem 25E

Chapter 3.8, Problem 27E

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Here is the optimal tableau for a standard Max problem. zx1 x2 x3 24 81 82 83 rhs 1 0 5 3 0 6 0 1 .3 7.5 0 - .1 .2 0 0 28 360 0 -8 522 0 2700 0 6 12 1 60 0 0 -1/15-3 1 1/15 -1/10 0 2 Using that the dual solution y = CBy B-1 and finding B = (B-¹)-¹ we find the original CBV and rhs b. The allowable increase for b₂ is If b₂ is increased by 3 then, using Dual Theorem, the new value for * is If c₂ is increased by 10, then the new value for optimal > is i.e. if no change to BV, then just a change to profit on selling product 2. The original coefficients c₁ = =☐ a and c4 = 5 If c4 is changed to 512, then (first adjusting other columns of row0 by adding Delta times row belonging to x4 or using B-matrix method to update row0) the new optimal value, after doing more simplex algorithm, for > is

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The equation of motion and the time at which the mass attains its extreme displacement from equilibrium position and find the position of the mass at this instant.

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